
\chapter{Arclength, Parameterized Curves, Ellipses} 

\section[Review]{Review of Parameterized Curves}

We first briefly review \index{parameterized curves} parameterized curves.  For a more thorough review,
reread the two chapters on this topic in the Math 113 notes.
% Math 124 notes.

\smallskip

A curve \index{parameterized by time} parameterized by time can be thought of as the motion of a particle
in the $xy$-plane whose $x$- and $y$-coordinates are each given by a
function of $t$: $x=x(t)$, $y=y(t)$.  Then $\dot{x}(t)=dx/dt$ is the
\index{horizontal velocity} horizontal velocity, $\dot{y}(t)=dy/dt$ is the vertical velocity, and
$dy/dx=\dot{y}(t)/\dot{x}(t)$ is the slope of the particle's path at the
particular time $t$.  In some situations we have other parameters besides
$t$, for example, an angle parameter $\theta$.  In that case we can still
compute the slope of the curve $dy/dx$ by taking the ratio of $dy/d\theta$
to $dx/d\theta$.

\smallskip

Some basic examples to understand are:

\smallskip

\noindent
1.~{\bf Straight line path from $P=(x_0,y_0)$ to $Q=(x_1,y_1)$.} If you
travel from $P$ to $Q$ at constant speed in time $T$, then your $x$- and
$y$-coordinates at time $t$ are: 
\[
x(t)=x_0+\bigl((x_1-x_0)/T\bigr)t, \text{ and }
y(t)=y_0+\bigl((y_1-y_0)/T\bigr)t\,.
\]
\smallskip
2.~{\bf Falling bodies (no air resistance).} If the object starts from
$(x_0,y_0)$ with \index{initial velocity} initial velocity of $v_0$ directed at an angle $\alpha$
above the horizontal ($\alpha$ is a negative angle if it's being thrown
down), then 
\[
  x(t)=x_0+(v_0\cos\alpha)t \text{ and }
y(t)=-\frac{1}{2}gt^2+(v_0\sin\alpha)t+y_0\,.
\]
\smallskip
3.~{\bf Circular motion.} If the particle is moving around the \index{circle} circle of
radius $r$ centered at the point $(h,k)$, and if we let $\theta$ denote the
angular position of the particle in radians measured counterclockwise from
the positive $x$-direction, then $x(\theta)=h+r\cos\theta$,
$y(\theta)=k+r\sin\theta$.  If $\theta$ is a known function of $t$, then we
can substitute that function here to get $x(t)$, $y(t)$.  If the particle's
angular position at time $t=0$ is $\theta_0$ and if it has {\bf constant}
angular velocity (in radians/sec) equal to $\pm\omega$ --- where we take
$\omega$ when it is moving counterclockwise and $-\omega$ when it is moving
clockwise --- then 
\[
   x(t)=h+r\cos(\theta_0\pm\omega t),
\text{ and }
  y(t)=k+r\sin(\theta_0\pm\omega t)\,. 
\]
Some important special cases are: 
\begin{enumerate}
\item%(1) 
$h=k=\theta_0=0$, so that $x(t)=r\cos(\omega t)$, $y(t)=r\sin(\omega t)$; 
\item%(2) 
$\theta_0=\pi/2$ (i.e.,
it starts from the positive $y$-direction) and the rotation is clockwise,
so that $x(t)=h+r\cos(\frac{\pi}{ 2}-\omega t)= h+r\sin(\omega t)$,
$y(t)=k+r\sin(\frac{\pi}{ 2}-\omega t)=k+r\cos(\omega t)$; 
\item%(3) 
clockwise rotation starting from the {\it negative} $y$-direction, in which case
$x(t)=h-r\sin(\omega t)$, $y(t)=k-r\cos(\omega t)$.
\end{enumerate}

\smallskip

\noindent
4.~{\bf \index{Combined circular motion} Combined circular motion with some other motion.} Here the particle
is whirling around a center which itself is moving.  In that case you use
the circular motion equations in the preceding paragraph, but with $h$ and
$k$ (the coordinates of the center) replaced by the functions
$x_{\text{center}}(t)$ and $y_{\text{center}}(t)$ which describe the motion of
the center.  For example, the {\bf trochoid} is the motion of a point $P$
which is stuck at a distance $a$ (where $a$ is between 0 and $r$) from the
center of a wheel (where we assume that at time $t=0$ the wheel passes over
the origin with $P$ located directly below the center at that instant).  In
that case the circular motion of $P$ around the center has radius $a$, and
from special case (3) in the preceding paragraph we obtain the equations
\[
x(t)=r\omega t-a\sin(\omega t),
\text{ and } 
y(t)=r-a\cos(\omega t)\,. 
\]
The {\bf \index{cycloid} cycloid} is the special case of this when $a=r$.  If the wheel is
traveling along at a known linear velocity (say, in m/sec), then the
angular velocity $\omega$ (in rad/sec) is obtained by dividing the linear
velocity by the radius of the wheel.  If you are given $\omega$ in units of
revolutions/sec, then multiply by $2\pi$ to get rad/sec.


\begin{example}
%{\bf Example 1.}
A wheel of radius 0.2 meters rolls without slipping along
the flat roof of a building at speed 6 m/sec.  At time $t=0$ it flies off
the roof.  The building 

\begin{rightindent}{3.0 truein}
\noindent
is 122.5 m high.  Let $P$ be the point on the rim of the wheel which is at the
top at the instant $t=0$ when the wheel leaves the roof.  Find parametric
equations for the motion of $P$.  Take the side of the building as the
$y$-axis and the ground as the $x$-axis.

When the wheel rolls along the roof, its \index{angular velocity} angular velocity is: (linear
velocity)$/$radius$=6/0.2=30$ rad/sec.  After leaving the roof it continues
spinning at the same rate.  So the point $P$ revolves clockwise around the
center of the wheel at $\omega=30$ rad/sec, starting above the center when
$t=0$.  This is special case (2) in the paragraph above on circular motion.
Thus, the motion of $P$ relative to the center has equations $0.2\sin(30t)$
and $0.2\cos(30t)$.  Meanwhile, the center of the wheel starts from
\end{rightindent}
\hfill
%-figure DA -----------------------------
\begin{psfigure}{2.0in}{2.75in}{figDA.ps}
%\putcrosshair{0,0}
\puttext{20., 20.}{\small Path of center}
\put(16,16){\vector(-1,-1){8}}
\puttext{-30., 35}{\small$P$}
\put(-30,29){\vector(1,-1){8}}
\end{psfigure}
%------------------------------

\noindent
$(0,122.7)$ with initial horizontal component of velocity 6 m/sec (and zero
initial vertical component of velocity).  So $x_{\text{center}}(t)=6t$,
$y_{\text{center}}(t)=-4.9t^2+122.7$.  Finally, we get the equations for
$P$ by adding its equations relative to the center to the equations for the
coordinates of the center: $$ x(t)=6t+0.2\sin(30t),\qquad
y(t)=-4.9t^2+122.7+0.2\cos(30t).  $$

\end{example}


\section{Ellipses}

An \index{ellipse} ellipse is obtained by stretching the circle of radius 1
by a factor of $a$ in one direction and by a factor of $b$ in the
perpendicular direction.  An ellipse has two axes of symmetry, the {\bf
\index{major axis} major axis}, having length $2a$, which joins the two farthest opposite
points on the circumference, and the {\bf \index{minor axis} minor axis}, having length $2b$,
which joins the two closest opposite points on the circumference.  The
half-length $a$ is called the {\bf semi-major axis}, and $b$ is called the
{\bf semi-minor axis}.  If we choose to locate the ellipse in the
$xy$-plane so that its center is at the origin, its major axis is in the
$x$-direction, and its minor axis is in the $y$-direction, then it has the
implicit equation $x^2\!/a^2+y^2\!/b^2=1$.  (The special case $a=b=r$ is
simply the circle of radius $r$, having equation $x^2+y^2=r^2$.)  In
parametric form, the ellipse can be described by the equations
$x(\theta)=a\cos\theta$, $y(\theta)=b\sin\theta$, where we go once around
the ellipse as $\theta$ goes from 0 to $2\pi$.  (One difference with a
circle, however: the point $P=(x(\theta),y(\theta))$ is {\it not} located
at an angle $\theta$ from the positive $x$-axis.  In fact, if $\alpha$ is
the angular position of $P$, then
$\tan\alpha=y(\theta)/x(\theta)=(b/a)\tan\theta$, which is not the same as
$\tan\theta$ when $a\ne b$; hence $\alpha$ is not the same as $\theta$.)

\smallskip
\begin{rightindent}{3.0 truein}
 
Another description of an \index{ellipse} ellipse is the locus of all points $P$ for which
the sum of the distance from $P$ to two fixed points is a constant.  Those
two points are called the {\bf foci} (the plural of {\it \index{focus of an ellipse} focus}).  If the
ellipse is situated in the standard way, with the major axis along the
$x$-axis, then both foci are on the $x$-axis, equidistant from the origin.
We let $(c,0)$ and $(-c,0)$ denote the coordinates of the foci.  See the
drawing at the right.

 
\end{rightindent}
\hfill
%-figure DB -----------------------------
\begin{psfigure}{2in}{1.75in}{figDB.ps}
%\putcrosshair{0,0}
\puttext{-50., -7.5}{\small$-a$}
\puttext{-27.5, -7.5}{\small$-c$}
\puttext{50., -7.5}{\small$a$}
\puttext{25., -7.5}{\small$c$}
\puttext{-5., -40.}{\small$-b$}
\puttext{-2.5, 40.}{\small$b$}

\end{psfigure}
%------------------------------

We can get an important relationship between $a$, $b$ and $c$ if we notice
that (1) the point $(a,0)$ on the ellipse is at a distance $a+c$ from the
farther focus and $a-c$ from the closer focus, so the sum of the distances
is $2a$; and (2) the point $(0,b)$ on the ellipse is at a distance
$\sqrt{b^2+c^2}$ from each of the two foci, so the sum of the 
distances is $2\sqrt{b^2+c^2}$.  Since these sums of distances must be the
same, it follows that $2a=2\sqrt{b^2+c^2}$.  Dividing by 2 and squaring
both sides, we conclude that $a^2=b^2+c^2$.  This relation enables us to
find one of the three constants $a$, $b$, $c$ if we know the other two.


\smallskip

The \index{{\bf eccentricity} of an ellipse} {\bf eccentricity} of an ellipse is the ratio $e=c/a$, i.e., the
fraction of the way from the center to the circumference that a focus is
located on the $x$-axis.  The eccentricity is always between 0 and 1: $0\le
e<1$.  If $e=0$, then $a=b$ and we have a circle.  The greater $e$, the
farther our ellipse is from looking like a circle.  Ellipses with $e$ near
1 are long and thin.

\begin{example}
%{\bf Example 2.} 
Find the equation of the ellipse of eccentricity 0.8 whose
foci are at $(0,\pm 60)$.

\smallskip
 
Here the foci are on the $y$-axis rather than the $x$-axis.  This means
that the major axis is in the $y$-direction.  The distance $c$ from the
center to a focus is 60, and $e=0.8=c/a$.  Hence, $a=c/e=60/0.8=75$.  Then
$b=\sqrt{a^2-c^2}=45$.  When the major axis is in the $y$-direction, the
equation of the ellipse is $x^2\!/b^2+y^2\!/a^2=1$, i.e., our answer is
$\disp
\frac{x^2}{ 2025}+\frac{y^2}{ 5625}=1.
$
\end{example}

\medskip

We now give a second way to describe an ellipse parametrically.
We first move the ellipse horizontally a distance $c$ to the right so that
it has a focus at the origin.  Then the other focus is at the point
$(2c,0)$.  (Sometimes it is more natural to put the origin at a focus of
the ellipse rather that at its center.  For example, the orbit of a planet
as it revolves around the sun is an ellipse with the sun at one focus.)
See the diagram on the next page.

\newpage



\begin{rightindent}{3.5in}
If $(x,y)$ are the coordinates of a point on the ellipse, then for our
parameter we use the angle $\theta$ that the line from the origin to $(x,y)$
makes with the $x$-axis.  (As mentioned above, the $\theta$-parameter in
our earlier \index{parametric equations} parametric equations does {\it not} have this simple
geometrical meaning.)  
Let $r=\sqrt{x^2+y^2}$ denote the distance of
the point from the origin, and let $d$ be its distance from the other
focus.  Since the sum of the distances of $(x,y)$ from the two foci is
equal to $2a$, it follows that $d=2a-r$.  On the other hand, by the
law of cosines, $d^2=r^2+(2c)^2-2r(2c)\cos\theta$.  Thus,
\index{ellipse parameterized from a focus} 
\end{rightindent}
\hfill
% Diagram showing ellipse parameterized from a focus.
%-figure DBb -----------------------------
%
\begin{psfigure}{2.5in}{2.5in}{figDBb.ps}
%\putcrosshair{0,0}
\puttext{25,-5}{\tiny$(2c,0)$}
\puttext{-10,5}{\small$\theta$}
\puttext{15,30}{\small$(x,y)$}
\puttext{-2,20}{\small$r$}
\end{psfigure}
%------------------------------



$$
      (2a-r)^2=r^2+4c^2-4rc\cos\theta.
$$  If we expand the square on the left,
then cancel the $r^2$ and divide through by 4, we get
$a^2-ar=c^2-rc\cos\theta,$ and so $$r=\frac{a^2-c^2}{ a-c\cos\theta}.$$
Since $x=r\cos\theta$ and $y=r\sin\theta$, we immediately obtain
formulas for $x$ and $y$ in terms of the parameter $\theta$:
$$
 x=\frac{(a^2-c^2)\cos\theta}{ a-c\cos\theta},\qquad
y=\frac{(a^2-c^2)\sin\theta}{ a-c\cos\theta},\qquad 0\le\theta\le 2\pi.
$$


\section{Arclength} 

Suppose we have a curve described parametrically by equations $x=x(t)$,
$y=y(t)$.  By the {\bf \index{arclength} arclength} of the curve between $t=a$ and $t=b$ we
mean the actual length of the path (distance traveled).  That is, if you
were to imagine the path made of string and then straighten out the string
and put it along the $x$-axis, this would be the length along the $x$-axis.
In what follows we shall use the notation $s(t)$ to mean the distance along
the path from some reference point (perhaps the beginning of the path) to
the point $(x(t),y(t))$.  Then the arclength of the path between $t=a$ and
$t=b$ is simply $s(b)-s(a)$.

\begin{center}
%-figure DC -----------------------------
\begin{psfigure}{4in}{1.75in}{figDC.ps}
%\putcrosshair{0,0}
\puttext{-20., -5.}{\small$t=b$}
\puttext{-57.5, -12.5}{\small$t=a$}
\end{psfigure}\\
%------------------------------
\end{center}

The derivative \index{$ds/dt$} $ds/dt$ --- the rate at which the particle is traveling
along its path --- is what we normally call its speed.  This is the length
of the velocity vector, i.e., the hypotenuse of the triangle made up of the
horizontal and vertical components of velocity.  Thus, 
$$
\frac{ds}{
dt}=\sqrt{\left(\frac{dx}{ dt}\right)^2+\left(\frac{dy}{ dt}\right)^2}.
$$ 
This
means that the distance function $s(t)$ is an antiderivative of the square
root on the right, and so we have
$$
\Bigl(\text{arclength~from~}t=a\text{~to~}t=b\Bigr)=s(b)-s(a)=\int_a^b\sqrt{
\left(\frac{dx}{ dt}\right)^2+\left(\frac{dy}{ dt}\right)^2}\>dt.
$$
In words, this says that {\it distance traveled is equal to the area under
the graph of speed}.

\begin{example}
%{\bf Example 3.} 
Find a definite integral for the distance traveled by the
point $P$ in Example 4.1 from when the wheel leaves the roof to when it hits
the ground.

\smallskip

The wheel leaves the roof at time $t=0$.  To find out when it hits the
ground, we observe that at that instant the center is at height 0.2 (the
radius).  Setting $0.2=y_{\text{center}}(t)=-4.9t^2+122.7$, we find that
$t=\sqrt{122.5/4.9}=5$ sec.  To find the speed, we compute the velocity
components $dx/dt=6+6\cos(30t)$, $dy/dt=-9.8t-6\sin(30t)$, and so the
distance traveled is
$$
\int_0^5\sqrt{(6+6\cos(30t))^2+(-9.8t-6\sin(30t))^2}\>dt.
$$ This integral
is too complicated to evaluate by the \index{antiderivative method} antiderivative method.  It must be
computed by ``numerical integration,'' i.e., an \index{approximation method} approximation method. In
the next section we shall present one such method, the \index{trapezoid rule} trapezoid rule.
\end{example}

\medskip

If your curve is given in terms of a parameter other than time, you can use
the same formula to determine the arclength --- there is nothing special
about the letter $t$ in the above formula.  For example, if $\theta$ is
your parameter, then the arclength of the curve between $\theta=a$ and
$\theta=b$ is 
$$
\int_a^b\sqrt{\left(\frac{dx}{ d\theta}\right)^2
    +\left(\frac{dy}{ d\theta} \right)^2}\>d\theta.
$$ 


\begin{example}
%{\bf Example 4.} 
Find the arclength of one arch of the cycloid, i.e.,
$x(\theta)=r\theta-r\sin\theta$, $y(\theta)=r-r\cos\theta$, between
$\theta=0$ and $\theta=2\pi$.  Express your answer in terms of $r$.  This
is the actual distance traveled by a pebble in the tread of the wheel each
time the wheel makes one revolution.  If the bicycle travels 1 mile, how
far does the pebble travel?

\smallskip

Here $dx/d\theta=r(1-\cos\theta)$ and $dy/d\theta=r\sin\theta$.  Before
taking the integral, it's helpful to simplify the sum of the squares of the
derivatives.  We do this by factoring out $r^2$ and then using the trig
identity $\sin^2+\cos^2=1$: 
\begin{align*}
\left(\frac{dx}{ d\theta}\right)^2+\left(\frac{dy}{ d\theta}\right)^2
&= r^2\Bigl((1-\cos\theta)^2+\sin^2\theta\Bigr)\\
&= r^2\Bigl(1-2\cos\theta +\cos^2\theta+\sin^2\theta\Bigr)
=r^2(2-2\cos\theta).
\end{align*} 
We must integrate the square root of this:
$\int_0^{2\pi}r\sqrt{2-2\cos\theta}
\,d\theta$.  It is not obvious how to do this integral by the antiderivative
method.  For example, a $u$-substitution of the form $u=2-2\cos\theta$
would {\bf not} work.  (Such a substitution would work if we had a
$\sin\theta$ term sitting beside the square root --- but unfortunately we
don't have that.)  Perhaps we should wait until the trapezoid rule, and be
satisfied with an approximate value.  However, there is a trick that will
make the integrand take a simpler form.  It uses the double angle formula
$\cos(2A)=1-2\sin^2(A)$.  (We will later find that this formula comes up in
integration problems in other situations --- see the section on integrating
powers of sine and cosine.)  If we replace $A$ by $\frac{1}{ 2}
\theta$ in this double angle formula, and then move the cosine to the right
and the $\sin^2$ to the left and multiply through by 2, we obtain:
$4\sin^2(\theta/2)=2-2
\cos(\theta)$.  Taking the square root of both sides, we get:
$\sqrt{2-2\cos\theta}=2\sin(\theta/2)$.  (Note: In general when we take
square roots we have to put a $\pm$ in front of the radical; however,
in this case, since we have assumed that $\theta$ is in the range between
$0$ and $2\pi$, and so $\theta/2$ is between $0$ and $\pi$, we know that
$\sin(\theta/2)$ is positive.)  Thus, the arclength of the cycloid
is 
\[
\int_0^{2\pi}2r\sin(\theta/2)\,d\theta= -2r\cdot\frac{1}{
1/2}\cos(\theta/2)\bigm|_0^{2\pi}=-4r(\cos\pi-\cos 0)=8r.
\] 
Thus, the
pebble travels a distance $8r$ during the time the wheel makes one
revolution, i.e., during the time the bicycle travels $2\pi r$.  Thus, the
pebble travels $8/2\pi$ times as far as the bicycle, and so when the
bicycle travels 1 mile the wheel travels $8/2\pi=4/\pi=1.27$ miles.

\end{example}

\medskip

Finally, we consider the case when a function is given in the explicit form
$y=f(x)$ between $x=a$ and $x=b$.  We can still use the same arclength
formula, because the explicit form of a curve is a special case of
parameterized equations: simply choose $x$ as your parameter, and take the
two equations $x=x$, $y=f(x)$.  Then $dx/dx=1$ and $dy/dx=f'(x)$.  Using
$x$ in place of $\theta$ in the above arclength formula, we obtain
\[
\Bigl(\text{arclength~of~the~curve~}y=f(x)\text{~from~}x=a\text{~to~}x=b\Bigr)=
\int_a^b\sqrt{1+(f'(x))^2}\,dx.
\]

\smallskip

In most cases --- even when dealing with simple basic functions --- these
arclength formulas lead to integrals that cannot be computed by the
\index{antiderivative method} antiderivative method, and we must use \index{numerical integration} numerical integration.

\begin{example}
%{\bf Example 5.} 
Find the length of one arch of the sine curve.

\smallskip

Here we want the arclength of $y=f(x)=\sin x$ from $x=0$ to $x=\pi$.  By
the previous formula, since $dy/dx=\cos x$ we have 
$$
\text{arclength}=
\int_0^\pi\sqrt{1+\cos^2x}\,dx.
$$
This integral cannot be computed by the antiderivative method.  In the
next section we will get an approximate value by the trapezoid rule.

\end{example}

\medskip


 
Sometimes instead of the arclength itself, one is interested in the {\it
rate} at which arclength is increasing --- i.e., the speed of the particle.
For example, given the horizontal component of
velocity, we might ask for $ds/dt$; or
given $ds/dt$, we might want to determine $dx/dt$ or $dy/dt$.  Here is a
slightly more complicated example of a related rate problem.


\begin{example} 
%{\bf Example 6.} 
\index{bug problem}
A bug is crawling up the $y=x^2$ parabola.  Meanwhile, the
sun is shining from the left from an angle of $30^\circ$ above the
horizontal, casting a shadow on the $x$-axis (see the figure).  When
the bug reaches the point $(1.5,2.25)$, its shadow is moving at the rate of
0.3 units/min.  Find the bug's horizontal and \index{vertical components} vertical components of
velocity and its speed.

\smallskip
\begin{rightindent}{3.0 truein}
Let $u(t)$ denote the location of the \index{bug's shadow} bug's shadow on the $x$-axis at time
$t$.  We know that $du/dt=0.3$, and we want to find $dx/dt$, $dy/dt$, and
$ds/dt=\sqrt{(dx/dt)^2+(dy/dt)^2}$.  We have a simple relationship between
$y$ and $x$, namely $y=x^2$, but we need to find a relationship between $u$
and $x$.  To do that we observe the geometry of the shadow.  Let $O$ be the
origin, $P$ be the point on the $x$-axis directly under the bug, and $Q$ be
the location of the bug's shadow on the $x$-axis.  Then
\end{rightindent}
\hfill
%-figure DD -----------------------------
\begin{psfigure}{2in}{2in}{figDD.ps}
%\putcrosshair{-22,0}
\puttext{-10., 20.}{\small bug}
\put(-10,17.5){\vector(-1,-1){10}}
\put(45,-12.5){\vector(-1,-3){5}}
\puttext{50., -12.5}{\small shadow}
\puttext{-50,-40}{\small$O$}
\puttext{-22,-40}{\small$P$}
\puttext{ 40,-40}{\small$Q$}
\end{psfigure}
%------------------------------

\noindent
$u=\overline{OP}+\overline{PQ}
=x+y\cot(30^\circ)=x+y\sqrt{3}=x+x^2\sqrt{3}$.  This gives us a relation
between $u$ and $x$.  Taking $d/dt$ of this relation, we obtain:
\[
0.3=\frac{du}{ dt}=\frac{d}{ dt}(x+x^2\sqrt{3})
=\frac{dx}{ dt}+2x\sqrt{3}\frac{dx}{ dt}=(1+2\cdot 1.5\sqrt{3})\frac{dx}{ dt}.
\]
Solving for $dx/dt$ gives us $dx/dt=0.3/(1+3\sqrt{3})=0.0484$ units/min.
Next, $dy/dt=(dy/dx)(dx/dt)=2x(dx/dt)=2\cdot 1.5\cdot 0.0484=0.1452$
units/min, and the speed is $\sqrt{0.0484^2+0.1452^2}=0.1531$ units/min.

\end{example}
% 
